The purpose of this supplement is to demonstrate that the motion of a body moving under the influence of Newton’s gravitational law—that is, the inverse square force law—moves in a conic section. In the text we dealt with circular orbits only, but some of the results, such as Kepler’s law on the relation between the period and the size of the orbit can be generalized to the case of elliptical orbits.
History. At this point it is a good idea to review some of the history given at the beginning of the text in addition to the relevant sections of the text (especially §4.1). Recall, for example, that one of the important observational points that was part of the Ptolemaic theory as well as being properly explained by the fact that planets move (to an excellent degree of approximation) in ellipses about the sun, is the appearance of retrograde motion of the planets, as in Figure4.3.1.
Figure 4.3.1.the apparent motion of the planets. The photograph shows the move-ments of Mercury, Venus, Mars, Jupiter, and Saturn. the diagram shows the paths traced by the planets as seen from the Earth, which the Ptolemaic theory tried to explain.
Methodology. Our approach in this supplement is to use the laws of conservation of energy and conservation of angular momentum. This means that we will focus on a combination of techniques using the basic laws of mechanics together with techniques from differential equations together with a couple of tricks. Other approaches, and in fact, Newton’s original approach were much more geometrical.10
The Kepler Problem. As in the text, we assume that the Sun (of mass M ) is so massive that it remains fixed at the origin and our planet (with mass m) revolves about the Sun according to Newton’s second law and moving in the field determined by Newton’s Law of Gravitation. Therefore, the Kepler problem is the following: Demonstrate that the solution r(t) of the equation
m¨r =GmM
r2 (4.3.3)
is a conic section. The procedure is to suppose that we have a solution, which we will refer to as an orbit, and to show it is a conic section. By reversing the argument, one can also show that any conic section, with a suitable time parametrization, is also a solution.
Energy and Angular Momentum. We saw in the paragraph in §4.3 of the text entitled Conservation of Energy and Escaping the Earth’s Grav-itational Field that any solution of equation (4.3.3) obeys the law of con-servation of energy; that is, the quantity
E = 1
2m˙r2−GmM
r (4.3.4)
called the energy of the solution, is constant in time.
The second fact that we will need is conservation of angular momen-tum, which was given in Exercise 20, §4.1. This states that the angular momentum of a solution is constant in time, namely the cross product
J = mr× ˙r (4.3.5)
is time independent for each orbit.
Orbits Lie in Planes. The first thing we observe is that any solution must lie in a plane. This is simply because of conservation of angular mo-mentum: since J is a constant and since J = mr× ˙r, it follows that r lies in the plane perpendicular to the (constant) vector J.
Introducing Polar Coordinates. From the above consideration, we can assume that our orbit lies in a plane, which we can take to be the usual
10The geometric approach, along with a lot of interesting history, is also emphasized in the little book Feynman’s Lost Lecture: The Motion of Planets Around the Sun by David L. Goodstein and Judith R. Goodstein, W.W. Norton and Co., New York.
xy-plane. Let us introduce polar coordinates (r, θ) in that plane as usual, so that the components of the position vector r are given by
r = (r cos θ, r sin θ).
Differentiating in t, we see that the velocity vector has components given by
˙r = ( ˙r cos θ− r ˙θ sin θ, ˙r sin θ + r ˙θ cos θ) and so one readily computes that
˙r2= ˙r2+ r2˙θ2. (4.3.6) Substituting (4.3.6) into conservation of energy gives
E = 1 2m
"
˙r2+ r2˙θ2#
−GmM
r (4.3.7)
Similarly, the angular momentum is computed by taking the cross product of r and ˙r:
J = m
i j k
r cos θ r sin θ 0
˙r cos θ− r ˙θ sin θ ˙r sin θ + r ˙θ cos θ 0
(4.3.8)
= mr2˙θ k (4.3.9)
Thus,
J = mr2˙θ (4.3.10)
is constant in time.
Kepler’s Second Law. We notice that the expression for J is closely related to the area element in polar coordinates, namely dA = 12r2dθ. In fact, measuring area from a given reference θ0, we get
dA dt = J
2m (4.3.11)
which is a constant. Thus, we get Kepler’s second law, namely that for an orbit, equal areas are swept out in equal times. Also note that this works for any central force motion law as it depends only on conservation of angular momentum.
Rewriting the Energy Equation. Next notice that the energy equa-tion (4.3.7) can be written, with ˙θ eliminated using (4.3.10), as
E = 1
2m ˙r2+ J2
2mr2 −GmM
r (4.3.12)
It will be convenient to seek a description of the orbit in polar form as r = r(θ).
A Trick. Solving differential equations to get explicit formulas often in-volves a combination of clever guesses, insight and luck; the situation with the Kepler problem is one of these situations. A trick that works is to intro-duce the new dependent variable u = 1/r (of course one has to watch out for the possibility that the orbit passes through the origin, in which case r would be zero.) But watch the nice things that happen with this choice.
First of all, by the chain rule, du dθ =−1
r2 dr dθ
and second, notice that, using the chain rule, the preceding equation and (4.3.10), and so from (4.3.12) we find that
E = J2
Notice the interesting way that only u and its derivative with respect to θ appears and that the denominators containing r have been eliminated.
This is the main purpose of doing the change of dependent variable from r to u.
Completing the Square. Next, we are going to eliminate the term linear in u in (4.3.14) by completing the square. Let α = GM m2/J2, so so that equation (4.3.15) becomes
dv
We make one more change of variables to w = 1
αv = 1 αu− 1 so that (4.3.16) becomes
dw dθ
2
+ w2= e2, (4.3.17)
where
e2=β2
α2 =2Em α2J2 − 1.
Solving the Equation. The solution of the equation (4.3.17) is readily verified to be
w = e cos(θ− θ0)
where θ0 is a constant of integration. In terms of the original variable r, this means that
r (e cos(θ− θ0) + 1) = 1
α. (4.3.18)
Orbits are Conics. Equation (4.3.18) in fact is the equation of a conic written in polar coordinates, where e is the eccentricity of the orbit, which determines its shape. The quantity l = 1/α determines its scale and θ0
its orientation relative to the xy-axes; it is also the angle of the closest approach to the origin.
In the case that 0≤ e < 1 (and E < 0) one has an ellipse of the form (x + ae)2
a2 +y2 b2 = 1 where
a = l
1− e2 =−GmM 2E and
b2= al =− J2 2mE.
The case e = 0 gives circular orbits. One also gets hyperbolic orbits if e > 1 and parabolic orbits in the special case when e = 1. Thus, as e ranges from 0 to 1, the ellipse gets a more and more elongated shape.
Kepler’s 3rd Law. From Kepler’s second law and the fact that the area of an ellipse is πab, one finds that the period T of an elliptical orbit is related to the semi-major axis a by
T 2π
2
= a3 GM
which is Kepler’s third law. Note that this reduces to what we saw in the text for circular orbits in the case of a circle (when a is the radius of the circle).